I always forget the value of $\tan(\frac{\pi}{8})$. Is there a quick way to derive it?

Yes! Use the half-angle formula: $\tan(\frac{\alpha}{2}) = \frac{1 - \cos \alpha}{\sin \alpha}$. For $\frac{\pi}{8}$, set $\alpha = \frac{\pi}{4}$. Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, the calculation becomes very straightforward. It is a great way to verify your memory under pressure.

JEE Main 2022 (28 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $\frac{π}{8}$ and $θ$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then $tan^{2} θ$ is equal to

Select Answer:

Visualized Solution

Visualize the Scenario

Two vertical poles: $A B$ and $P Q$ .
Distance between feet $B$ and $Q = 160$ m.

Identify the Midpoint $C$

$C$ is the midpoint of $B Q$ .
$B C = C Q = \frac{160}{2} = 80$ m.

Define Pole Heights

Let height of $A B = h$ .
Then, height of $P Q = 2 h$ (given).

Angles of Elevation

Angles of elevation from $C$ :
To $A$ : $θ$
To $P$ : $\frac{π}{8}$

Analyze $△ A B C$

In right $△ A B C$ :
$tan θ = \frac{A B}{B C} = \frac{h}{80}$
$⟹ h = 80 tan θ$ --- (1)

Analyze $△ P QC$

In right $△ P QC$ :
$tan \frac{π}{8} = \frac{P Q}{C Q} = \frac{2 h}{80}$
$⟹ tan \frac{π}{8} = \frac{h}{40}$
$⟹ h = 40 tan \frac{π}{8}$ --- (2)

Equate the Heights

Equating $h$ from (1) and (2):
$80 tan θ = 40 tan \frac{π}{8}$

Relate $tan θ$ and $tan \frac{π}{8}$

Divide by 40:
$2 tan θ = tan \frac{π}{8}$
$tan θ = \frac{1}{2} tan \frac{π}{8}$

Recall $tan \frac{π}{8}$ Value

Standard trigonometric value:
$tan \frac{π}{8} = 2 - 1$

Substitute the Value

Substitute this value:
$tan θ = \frac{2 - 1}{2}$

Calculate $tan^{2} θ$

Square both sides to find $tan^{2} θ$ :
$tan^{2} θ = (\frac{2 - 1}{2})^{2}$
$tan^{2} θ = \frac{( 2 - 1 ) ^{2}}{2 ^{2}}$

Expand the Numerator

Expand the numerator using $(a - b)^{2} = a^{2} + b^{2} - 2 ab$ :
$tan^{2} θ = \frac{( 2 ) ^{2} + ( 1 ) ^{2} - 2 ( 2 ) ( 1 )}{4}$
$tan^{2} θ = \frac{2 + 1 - 2 2}{4}$

Final Simplification

Final simplification:
$tan^{2} θ = \frac{3 - 2 2}{4}$
Correct Option: (3)

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine two vertical poles,

A B

and

P Q

, standing on a flat plain separated by a distance of

160

meters. A point

C

is located at the midpoint of the base line

B Q

Because

C

is the midpoint, it divides the

160

meter gap into two equal segments:

B C = 80 meters

C Q = 80 meters

The Trigonometric Bridge

Let the height of the first pole

A B

h

. Given that the height of the second pole

P Q

is twice that of the first, we have

P Q = 2 h

From point

C

, the angle of elevation to the top of pole

A

θ

, and the angle of elevation to the top of pole

P

\frac{π}{8}

. We now consider two right-angled triangles:

△ A B C

and

△ P QC

△ A B C

, the relationship is:

tan θ = \frac{A B}{B C} = \frac{h}{80} \Rightarrow h = 80 tan θ

△ P QC

, the relationship is:

tan (\frac{π}{8}) = \frac{P Q}{C Q} = \frac{2 h}{80} = \frac{h}{40} \Rightarrow h = 40 tan (\frac{π}{8})

The Algebraic Resolution

By equating the two expressions for

h

, we bridge the two triangles:

80 tan θ = 40 tan (\frac{π}{8})

Dividing both sides by

40

, we obtain:

2 tan θ = tan (\frac{π}{8}) \Rightarrow tan θ = \frac{1}{2} tan (\frac{π}{8})

Using the known value

tan (\frac{π}{8}) = 2 - 1

, we substitute to find:

tan θ = \frac{2 - 1}{2}

Final Calculation

To find

tan^{2} θ

, we square the expression:

tan^{2} θ = (\frac{2 - 1}{2})^{2}

Expanding the numerator using the identity

(a - b)^{2} = a^{2} + b^{2} - 2 ab

tan^{2} θ = \frac{( 2 ) ^{2} + 1 ^{2} - 2 ( 2 ) ( 1 )}{4} = \frac{2 + 1 - 2 2}{4}

The final result is:

tan^{2} θ = \frac{3 - 2 2}{4}

Common Traps & FAQs

Why can't I use the Law of Sines here? Isn't it more general?

You absolutely could, but in JEE Advanced, time is your currency. Using the Law of Sines on right-angled triangles is like using a sledgehammer to crack a nut. The tangent function is the most direct, elegant, and error-proof path. Stick to the simplest tool that gets the job done.

I always forget the value of

tan (\frac{π}{8})

. Is there a quick way to derive it?

Yes! Use the half-angle formula:

tan (\frac{α}{2}) = \frac{1 - c o s α}{s i n α}

. For

\frac{π}{8}

, set

α = \frac{π}{4}

. Since

cos (\frac{π}{4}) = \frac{1}{2}

and

sin (\frac{π}{4}) = \frac{1}{2}

, the calculation becomes very straightforward. It is a great way to verify your memory under pressure.

Wait, why do we square the result at the very end? Is it possible I misread the question?

This is the classic 'JEE Trap.' Always underline what the question asks for. It asks for

tan^{2} θ

, not

tan θ

. It is a test of your attention to detail, not just your math skills. Always perform a final sanity check on the target variable before circling your answer.

Select Answer:

Visualized Solution

Visualize the Scenario

Identify the Midpoint $C$

Define Pole Heights

Angles of Elevation

Analyze $△ A B C$

Analyze $△ P QC$

Equate the Heights

Relate $tan θ$ and $tan \frac{π}{8}$

Recall $tan \frac{π}{8}$ Value

Substitute the Value

Calculate $tan^{2} θ$

Expand the Numerator

Final Simplification

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Resolution

Final Calculation

Similar Questions

A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $α$ and $β$ at $A$ and $B$ respectively. $A B$ subtends an angle $γ$ at $Q$ . Find the height of the pole.

A pole stands vertically inside a triangular park $A B C$ . Let the angle of elevation of the top of the pole from each corner of the park be $\frac{π}{3}$ . If the radius of the circumcircle of $Δ A B C$ is 2, then the height of the pole is equal to:

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$ , then the height of $P$ (in $m$ ) above the line $A C$ is :

Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $tan^{- 1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

Select Answer:

Visualized Solution

Visualize the Scenario

Identify the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Define Pole Heights

Angles of Elevation

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PQC

Equate the Heights

Relate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan8π​

Recall .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan8π​ Value

Substitute the Value

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2θ

Expand the Numerator

Final Simplification

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Resolution

Final Calculation

Similar Questions

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

Identify the Midpoint $C$

Analyze $△ A B C$

Analyze $△ P QC$

Relate $tan θ$ and $tan \frac{π}{8}$

Recall $tan \frac{π}{8}$ Value

Calculate $tan^{2} θ$